Y~Erlang(n,1) , how to solve E(|Y-n|))?
I think that we have to divide integral by 2 parts, one with y>n, other one with y<n. But I couldn't progress anymore. someone know about this?
2026-04-08 14:29:52.1775658592
mathematical statistic question
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Consider the distribution of $Z_j=Y_j-1$, where $Y_j \sim Exponential(1)$, and $Y \stackrel{iid}{=} \sum_j Y_j$, (iid) then $Y-n = \sum_{j=1}^{n}Z_j$.
Derive the distribution of $Z_j, \ f_{Z_j}(z_j) = e^{-z+1}, z \geq -1$, and find its expectation. Then $\mathbf{E}[Y-n]=\sum_j\mathbf{E}[Z_j]$, because $Z_j$ are also iid.